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Excerpt from Chapter 4.2:

Let us now illustrate the energy principle with a more complicated problem, the screw jack shown in Fig. 4–5. A handle 20 inches long is used to turn the screw, which has 10 threads to the inch. We would like to know how much force would be needed at the handle to lift one ton (2000 pounds). If we want to lift the ton one inch, say, then we must turn the handle around ten times. When it goes around once it goes approximately 126 inches. The handle must thus travel 1260 inches, and if we used various pulleys, etc., we would be lifting our one ton with an unknown smaller weight W applied to the end of the handle. So we find out that W is about 1.6 pounds. This is a result of the conservation of energy.

Figure 4.5:

Figure 4.5

I understand Feynman mathematically, 1.6 pounds * 1260 inches ~ 2000 pounds 1 inch up, but I don't understand his explanation visually.

  1. Why does the handle go 126 inches when it traverses the jack once?
  2. How exactly would we use the pulleys and the weight $W$? It's hard for me to picture this. I can only imagine a person pushing the handle with their hands.
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You have a good answer from ad2004 explaining that the handle's 20 inch radius equals an approximately 126 inch circumference. As for the weight and pulleys question, it is to demonstrate that a 1.6 pound force would move the handle. Suppose that the 20 inch horizontal handle was replaced with a horizontal 20 inch radius pulley. Now attach a string to the edge of the pulley, wrap it around the pulley 10 times, then run it over a fixed vertical pulley, and attach a 1.6 pound weight to the string. Ideally, the weight puts 1.6 pounds tension on the string to rotate the horizontal pulley 10 times as the weight lowers. This is one way of doing it, I hope it will help your understanding of the question.

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  • $\begingroup$ Nice, this was helpful. Just for the sake of the argument though, I think Feynman was referring to use pulleys alongisde the handle, not in its place: "We would be lifting our one ton with an unknown smaller weight W applied to the end of the handle". $\endgroup$ Nov 28 '20 at 17:06
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The following may be useful. One turn of the handle corresponds to $2 \pi R$ where $R$ is the length of the handle, $20$ inches. $2 \pi \times 20=126$ inches, the distance traveled by the end of the handle.

I hope this helps.

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  • $\begingroup$ Thanks, this was useful! Do you also know how we meant to use the pulleys and the weights to move the handle? $\endgroup$ Nov 27 '20 at 22:26

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